The energy of a photon can be calculated using the relationship E = hf where h is Planck's constant 6.626 x 10^-34 m^3 kg /s and f is the frequency in s^-1 (Hz). The energy of this photon is 3.13 x 10^-31 J.
Here is your equation: E=hf where h is "Planck's constant" (6.626 x 10^-34 J x sec) and f is the frequency.
E=hf
E=(6.626 x 10^-34 J x sec) x (4.8 x 10^15 Hz)
E=3.2 x 10-18 J
If I counted correctly (and I probabely didn't) your asking the energy of a photon whose frequency is 5.8*10^40 (a much neater convention).
Its a simple problem. E=hf
h=6.6*10^(-34)
So E=(6.6*10^(-34))(5.8*10^40)
I'll let you work it out for yourself.
The energy of a photon is given by the relation
E = hf
In this equation, h is called "Planck's constant" and has a value of 6.26 x 10^-34 Joules*sec. The frequency of the photon is f, and therefore the energy of the photon follows directly from this relation. The energy of a photon whose frequency is 5.0 * 10^14 Hz is
E = (6.26 * 10^-34 Joules*sec) * (5.0 * 10^14 Hz)= 3.13 * 10 ^-19 Joules
In particle physics, which deals with numbers on a much smaller scale than classical physics, the units of "electron-volts (eV)" are used. One electron-volt is the energy that an electron gains after traveling from a point in space to another point in space which is 1 Volt higher. Or as a physicist would say, "An electron-volt is the energy obtained by an electron after traveling through a 1 volt potential difference of an electric field."
Since a photon is a particle and belongs to the realm of particle physics, its energy is best described in terms of electron-volts, in which case
1 electron-Volt = 1eV = 1.602 *10^-19 Joules
And
E = hf = 3.13 * 10^-19 Joules = 1.95 eV
The energy of a photon whose frequency is 5.8 x 1014 Hz is: 2.4322643592E-11 electron volts.
Photon energy = (Planck's Konstant) x (frequency) =
(6.63 x 10-34 joule-second) x (5 x 1014 per second) = 3.315 x 10-19 joule
E=5.8*10^14*6.626*10^-34=3.84*10^-19
The energy is 2,3160 eV.
The energy E of a photon is E= h x f , where f is the frequency, and h is Planck's constant.h=6.63e-34 [aka 6.63 x 10^-34] J s (Joule-seconds)The frequency f of a photon is related to its wavelength byf=speed of light/wavelengthSpeed of light c=3e8 m/s.From this you can calculate the energy of a 600nm photon, and then the number of those photons required to get 4e-17 Joules.Your answer should be between 50 and 500, but you can do the math.
A microwave signal at 50 GHz has waves that are 10,000 times as long as a visible signal at yellow (600 nm) has. Therefore the yellow photon carries 10,000 times as much energy as the 50 GHz photon does.
50 hertz
It's "hertz", not "hetz". The number of hertz is the frequency. For example, if the current has a frequency of 50 hertz... well, that's the frequency. Hertz is equivalent to cycles/second, and it is sometimes written that way. So, 50 hertz could also be written as 50 cycles/second.
Wavelength= 10 mm. Frequency= 5.0 hertz. Speed= 50 mm/second (wavelength x frequency)
For a thermal radiation source, the peak of the blackbody radiation curve is at a photon energy 2.8 times the temperature in electron-volts. The temperature in electron-volts is 1/11,600 times the temperature in Kelvin. Use E = hv to convert from the photon energy (E) to photon frequency, using Plank's constant h. Use v = c/(lambda) to convert from the photon frequency to the wavelength. The result: these hot plasmas radiate X-rays, and the peak wavelength is about 50 Angstroms, i.e. 5 nm.
The energy E of a photon is E= h x f , where f is the frequency, and h is Planck's constant.h=6.63e-34 [aka 6.63 x 10^-34] J s (Joule-seconds)The frequency f of a photon is related to its wavelength byf=speed of light/wavelengthSpeed of light c=3e8 m/s.From this you can calculate the energy of a 600nm photon, and then the number of those photons required to get 4e-17 Joules.Your answer should be between 50 and 500, but you can do the math.
A microwave signal at 50 GHz has waves that are 10,000 times as long as a visible signal at yellow (600 nm) has. Therefore the yellow photon carries 10,000 times as much energy as the 50 GHz photon does.
50-100 USD
Nope.-- Visible light has frequencies roughly from 4 x 1014 Hz to 8 x 1014 Hz ... arange from lowest to highest of 4 x 1014 Hz, or one octave.-- Let's say that the lowest frequency in the total spectrum is 50 Hz (Europeanpower-line frequency) and the highest is 1019 Hz (lowest freq gamma rays) ...a total range from lowest to highest of (1019 - 50) Hz, or about 57 octaves.-- On a linear scale, of the total number of Hz in the spectrum, visible lightcovers about 0.004 percent of them.-- On a logarithmic scale, visible light covers 1 octave out of 57 octaves,or 1.74 percent of the spectrum.
Period = reciprocal of frequency ( 1 / frequency ) = 1/50 = 0.02 second = 20 milliseconds
A photon or light particle that comes from the sun has a certain amount of energy. Blue photons for example have an energy of 0.4 aJ. Red photons have half that energy. If a photovoltaic cell is built, a certain energy band gap is chosen. If it is too large, a photon will not be absorbed, if it is too small, part of the energy is lost. For instance if a photon of 0.4 aJ hits a solar cell of 0.3 aJ band gap, it will be absorbed and create an electron with energy 0.3 aJ. This gives an efficiency of 75%. If a 0.2 aJ photon hits this solar cell it will not be absorbed. Of the total energy of 0.6 aJ hitting the solar cell only 0.3 aJ is absorbed, a total efficiency of 50%. The highest efficiencies are reach with solar cells that contain several layers with different (decreasing) band gaps.
If a bulb has 50 Hz frequency and it's supply is 60 Hz frequency, it will still glow, despite the allowance of 10 HZ frequency.
50 Hz as per the info available from their official website
Frequency = 1/0.02 = 50 cycles per second or 50 Hertz
frequency of AC current in egyptis 50 c/s
50 hertz